Relative to an origin O, the position vectors of points A and B are given by
OA = 3pi + 4j + p^2k, and
OB = -pi + -1j + p^2k
Find the values of p for which angle AOB is 90°
So here's my work so far:
If angle = 90° then OA.OB = 0
OA.OB = p^4 - 3p^2 - 4
p^4 - 3p^2 - 4 = 0
Let x = p^2
x^2 - 3x - 4 = 0
x = 4 or -1
What do I do now?
Thanks for the help!
remember you let y = p^2
so p^2 = 4 ----> p = ± 2
or
p^2 = -1 , which has no real solution.
so p = ± 2
Wait, but doesn't the question say 'values'?
Oh wait, just realized how dumb that was! Had a brain fart, thanks for clearing things up!
To find the values of p for which the angle AOB is 90°, you have correctly set up the equation OA.OB = 0 and simplified it to p^4 - 3p^2 - 4 = 0.
Now, you have obtained the equation x^2 - 3x - 4 = 0 by substituting x = p^2. To continue solving this equation, you can factorize it or use the quadratic formula.
Factorizing the equation:
(x - 4)(x + 1) = 0
Setting each factor equal to zero:
x - 4 = 0 or x + 1 = 0
Solving for x:
x = 4 or x = -1
Since x = p^2, you can substitute back to find the possible values of p:
For x = 4:
p^2 = 4
Taking the square root:
p = ± 2
For x = -1:
p^2 = -1
Here, p^2 cannot be negative, as it is not a real number. Therefore, there are two possible values of p for which the angle AOB is 90°: p = 2 and p = -2.
Thus, the values of p for which angle AOB is 90° are p = ±2.